Prove that, if $u_n$ converges to $u$ strongly in $L^1(\mathbb R)$, then $$\frac{u_n}{1+u^2_n} \longrightarrow \frac{u}{1+u^2}\quad \text{ strongly in }L^1(\mathbb R).$$
My solution:
Put $h(s):=\frac{s}{1+s^2}$. It is differentiable and $|h'(s)|\leq 1$, then it is Lipschitz with constant equal to $1$. So we have
$$\bigg|\frac{u_n}{1+u^2_n} - \frac{u}{1+u^2}\bigg| \leq \big|u_n - u \big|$$
obtaining that
$$\int_\mathbb{R}\, \big|\frac{u_n}{1+u^2_n} - \frac{u}{1+u^2}\big|\, dx \leq \int_{\mathbb{R}} \big|u_n - u \big| \, dx \longrightarrow 0.$$
I simply used the particular form of the function. I did not use any convergence theorem of measure theory (e.g. dominated convergence). I would also like to see a solution that makes use of these tools. For example, I initially tried to solve the exercise as follows (which only works on $ L^1 (A) $ with $ | A | <\infty) .$
Since $ L^1 $ is a metric space I know that a sequence $ f_n $ converges to $ f $ if and only if every of its subsequences admit a convergent subsequence to $ f $. So let $ f_n = \frac{u_n}{1+u^2_n} $, $\, f=\frac{u}{1+u^2}$ and $ f_{n_k} $ be a subsequence. I know there is a sub-sequence of $ u_{n_k} $ that is almost everywhere convergent to $u$. So $ f_{n_{k_j}} $ also converges almost everywhere to $f$ and moreover $ | f_{n_{k_j}} | \leq 1 $. At this point I would like to use dominated convergence but I can not because $ h(x) \equiv 1 \notin L^1(\mathbb R) $. Is there anyone able to solve this exercise using measure theory tools? Maybe using almost uniform convergence? In general, I am interested in more solutions. The reason why I ask this is that I want to understand with greater generality how to show convergence in the case of space with not finite measure.
There is stronger (than usual) version of DCT: if $|f_n| \leq |g_n|$almost everywhere for each $n$,$f_n \to f$ almost everywhere and $g_n \to g$ in $L^{1}$ then $\int |f_n-f| \to 0$. I am sure you can you can find this on MSE, but you can also try to prove it yourself by splitting $\int |f_n-f| \to 0$ into integral over $\{|f_n| >g\}$ and its complement. Your result follows immediately from this (by going to almost everywhere convergent subsequences) since $|\frac {u_n} {1+u_n^{2}}| \leq |u_n|$.